cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A345101 Irregular triangle T(n, k) read by rows, n >= 0, k = 1..A000119(n); the n-th row contains the numbers m such that A022290(m) = n, in increasing order.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 8, 7, 9, 10, 11, 12, 16, 13, 17, 14, 18, 15, 19, 20, 21, 22, 24, 32, 23, 25, 33, 26, 34, 27, 28, 35, 36, 29, 37, 30, 38, 40, 31, 39, 41, 42, 43, 44, 48, 64, 45, 49, 65, 46, 50, 66, 47, 51, 52, 67, 68, 53, 69, 54, 56, 70, 72, 55, 57, 71, 73
Offset: 0

Views

Author

Rémy Sigrist, Jun 08 2021

Keywords

Comments

When interpreted as a flat sequence, yields a permutation of the nonnegative integers.

Examples

			Triangle begins:
     0    [0]
     1    [1]
     2    [2]
     3    [3, 4]
     4    [5]
     5    [6, 8]
     6    [7, 9]
     7    [10]
     8    [11, 12, 16]
     9    [13, 17]
    10    [14, 18]
    11    [15, 19, 20]
    12    [21]
		

Crossrefs

Cf. A000119 (row lengths), A003714, A003754, A022290.

Programs

  • PARI
    See Links section.

Formula

T(n, 1) = A003754(n+1).
T(n, A000119(n)) = A003714(n).

A357180 First run-length of the n-th composition in standard order.

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 1, 3, 1, 1, 2, 1, 1, 1, 2, 4, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 2, 3, 5, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 4, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 2
Offset: 0

Views

Author

Gus Wiseman, Sep 24 2022

Keywords

Comments

A composition of n is a finite sequence of positive integers summing to n. The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

Examples

			Composition 87 in standard order is (2,2,1,1,1), so a(87) = 2.
		

Crossrefs

See link for sequences related to standard compositions.
For parts instead of run-lengths we have A065120, last A001511.
The version for Heinz numbers of partitions is A067029, last A071178.
This is the first part of row n of A333769.
For minimal instead of first we have A357138, maximal A357137.
The last instead of first run-length is A357181.
A051903 gives maximal part in prime signature.
A061395 gives maximal prime index.
A124767 counts runs in standard compositions.
A286470 gives maximal difference of prime indices.
A333766 gives maximal part of standard compositions, minimal A333768.
A353847 ranks run-sums of standard compositions.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Table[If[n==0,0,First[Length/@Split[stc[n]]]],{n,0,100}]

A361676 a(n) is the greatest k such that n appears in the k-th row of triangle A361644.

Original entry on oeis.org

0, 1, 2, 3, 5, 5, 6, 7, 11, 10, 10, 11, 13, 13, 14, 15, 23, 22, 21, 21, 21, 21, 22, 23, 27, 26, 26, 27, 29, 29, 30, 31, 47, 46, 45, 45, 43, 42, 42, 43, 43, 42, 42, 43, 45, 45, 46, 47, 55, 54, 53, 53, 53, 53, 54, 55, 59, 58, 58, 59, 61, 61, 62, 63, 95, 94, 93
Offset: 0

Views

Author

Rémy Sigrist, Mar 20 2023

Keywords

Comments

All terms belong to A003754.
To compute a(n): consider the run lengths in the binary expansion of n (i.e. the n-th row of A101211) and replace from left to right each value v > 1 at even index with (1, v-1); at the end, there remain the run lengths in the binary expansion of a(n).
See A361645 for the least k's.

Examples

			The first terms, in decimal and in binary, are:
  n   a(n)  bin(n)  bin(a(n))
  --  ----  ------  ---------
   0     0       0          0
   1     1       1          1
   2     2      10         10
   3     3      11         11
   4     5     100        101
   5     5     101        101
   6     6     110        110
   7     7     111        111
   8    11    1000       1011
   9    10    1001       1010
  10    10    1010       1010
  11    11    1011       1011
  12    13    1100       1101
  13    13    1101       1101
  14    14    1110       1110
  15    15    1111       1111
		

Crossrefs

Programs

  • PARI
    See Links section.

Formula

a(n) >= n with equality iff n belongs to A003754.
a(n) >= A361645(n) with equality iff n belongs to A000975.

A361756 Irregular triangle T(n, k), n >= 0, k = 1..A361757(n), read by rows; the n-th row lists the numbers k such that the Fibonacci numbers that appear in the dual Zeckendorf representation of k also appear in that of n.

Original entry on oeis.org

0, 0, 1, 0, 2, 0, 1, 2, 3, 0, 1, 4, 0, 2, 5, 0, 1, 2, 3, 4, 5, 6, 0, 2, 7, 0, 1, 2, 3, 7, 8, 0, 1, 4, 9, 0, 2, 5, 7, 10, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 4, 12, 0, 2, 5, 13, 0, 1, 2, 3, 4, 5, 6, 12, 13, 14, 0, 2, 7, 15, 0, 1, 2, 3, 7, 8, 15, 16
Offset: 0

Views

Author

Rémy Sigrist, Mar 23 2023

Keywords

Comments

In other words, the n-th row lists the numbers k such that A003754(1+n) AND A003754(1+k) = A003754(1+k) (where AND denotes the bitwise AND operator).
The dual Zeckendorf representation is also known as the lazy Fibonacci representation (see A356771 for further details).

Examples

			Triangle T(n, k) begins:
  n   n-th row
  --  -------------------------------------
   0  0
   1  0, 1
   2  0, 2
   3  0, 1, 2, 3
   4  0, 1, 4
   5  0, 2, 5
   6  0, 1, 2, 3, 4, 5, 6
   7  0, 2, 7
   8  0, 1, 2, 3, 7, 8
   9  0, 1, 4, 9
  10  0, 2, 5, 7, 10
  11  0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11
  12  0, 1, 4, 12
		

Crossrefs

See A361755 for a similar sequence.

Programs

  • PARI
    See Links section.

Formula

T(n, 1) = 0.
T(n, 2) = A003842(n - 1) for any n > 0.
T(n, A361757(n)) = n.

A345291 a(n) is the least k >= 0 such that A345290(k) = n.

Original entry on oeis.org

0, 3, 2, 15, 9, 8, 11, 10, 63, 37, 36, 39, 33, 32, 35, 34, 47, 41, 40, 43, 42, 255, 149, 148, 151, 145, 144, 147, 146, 159, 133, 132, 135, 129, 128, 131, 130, 143, 137, 136, 139, 138, 191, 165, 164, 167, 161, 160, 163, 162, 175, 169, 168, 171, 170, 1023, 597
Offset: 0

Views

Author

Rémy Sigrist, Jun 13 2021

Keywords

Comments

The binary plot of the sequence has interesting features (see Illustration in Links section).

Examples

			We have:
  n         | 0   1  2  3   4   5   6   7  8  9  10  11  12  13  14  15  16  17
  ----------+------------------------------------------------------------------
  A345290(n)| 0  -1  2  1  -3  -4  -1  -2  5  4   7   6   2   1   4   3  -8  -9
- so a(0) = 0,
     a(2) = 2,
     a(1) = 3,
     a(5) = 8,
     a(4) = 9,
     a(7) = 10,
     a(6) = 11,
     a(3) = 15.
		

Crossrefs

Programs

  • PARI
    \\ See Links section.

A357138 Minimal run-length of the n-th composition in standard order; a(0) = 0.

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 1, 3, 1, 1, 2, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 0

Views

Author

Gus Wiseman, Sep 18 2022

Keywords

Comments

A composition of n is a finite sequence of positive integers summing to n. The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

Examples

			Composition 92 in standard order is (2,1,1,3), so a(92) = 1.
		

Crossrefs

See link for more sequences related to standard compositions.
The version for Heinz numbers of partitions is A051904, for parts A055396.
For parts instead of run-length we have A333768, maximal A333766.
The opposite (maximal) version is A357137.
For first instead of minimal we have A357180, last A357181.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Table[If[n==0,0,Min[Length/@Split[stc[n]]]],{n,0,100}]

A357181 Last run-length of the n-th composition in standard order.

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 1, 3, 1, 1, 2, 2, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 2, 2, 1, 1, 1, 5, 1, 1, 1, 2, 2, 1, 1, 3, 1, 1, 3, 2, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 2, 2, 1, 1, 1, 6, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 2, 2, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1
Offset: 0

Views

Author

Gus Wiseman, Sep 24 2022

Keywords

Comments

A composition of n is a finite sequence of positive integers summing to n. The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

Examples

			Composition 87 in standard order is (2,2,1,1,1), so a(87) = 3.
		

Crossrefs

See link for sequences related to standard compositions.
For parts instead of run-lengths we have A001511, first A065120.
For Heinz numbers of partitions we have A071178, first A067029.
This is the last part of row n of A333769.
For maximal instead of last we have A357137, minimal A357138.
The first instead of last run-length is A357180.
A051903 gives maximal part of prime signature.
A061395 gives maximal prime index.
A124767 counts runs in standard compositions.
A286470 gives maximal difference of prime indices.
A333766 gives maximal part of standard composition, minimal A333768.
A353847 ranks run-sums of standard compositions.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Table[If[n==0,0,Last[Length/@Split[stc[n]]]],{n,0,100}]

A374395 a(n) is the first term in the n-th row of A374394.

Original entry on oeis.org

0, 0, 0, 1, 0, 2, 2, 0, 1, 3, 3, 4, 0, 2, 2, 5, 6, 5, 7, 7, 0, 1, 3, 3, 4, 8, 10, 10, 8, 9, 11, 11, 12, 0, 2, 2, 5, 6, 5, 7, 7, 13, 14, 16, 16, 17, 13, 15, 15, 18, 19, 18, 20, 20, 0, 1, 3, 3, 4, 8, 10, 10, 8, 9, 11, 11, 12, 21, 23, 23, 26, 27, 26, 28, 28, 21
Offset: 0

Views

Author

Rémy Sigrist, Jul 10 2024

Keywords

Comments

a(n) is the least number z >= 0 such that the Zeckendorf representations of z and n-z have no common Fibonacci numbers and when combined together correspond to the lazy Fibonacci representation of n.

Crossrefs

Programs

  • PARI
    \\ See Links section.

Formula

a(n) = A022290(A374355(1+A003754(n)), k).
a(n) = n - A374396(n).

A374396 a(n) is the last term in the n-th row of A374394.

Original entry on oeis.org

0, 1, 2, 2, 4, 3, 4, 7, 7, 6, 7, 7, 12, 11, 12, 10, 10, 12, 11, 12, 20, 20, 19, 20, 20, 17, 16, 17, 20, 20, 19, 20, 20, 33, 32, 33, 31, 31, 33, 32, 33, 28, 28, 27, 28, 28, 33, 32, 33, 31, 31, 33, 32, 33, 54, 54, 53, 54, 54, 51, 50, 51, 54, 54, 53, 54, 54, 46
Offset: 0

Views

Author

Rémy Sigrist, Jul 10 2024

Keywords

Comments

a(n) is the greatest number z <= n such that the Zeckendorf representations of z and n-z have no common Fibonacci numbers and when combined together correspond to the lazy Fibonacci representation of n.

Crossrefs

Programs

  • PARI
    \\ See Links section.

Formula

a(n) = A022290(A374356(1+A003754(n)), k).
a(n) = n - A374395(n).

A375272 The number of factors of n of the form p^Fibonacci(k), where p is a prime and k >= 2, when the factorization is uniquely done using the dual Zeckendorf representation of the exponents in the prime factorization of n.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 2, 2, 1, 2, 2, 3, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 2, 3, 2, 2, 1, 3, 1, 2, 2, 3, 2, 3, 1, 2, 2, 3, 1, 3, 1, 2, 2, 2, 2, 3, 1, 3, 2, 2, 1, 3, 2, 2, 2
Offset: 1

Views

Author

Amiram Eldar, Aug 09 2024

Keywords

Comments

First differs from A086435 at n = 36. Differs from A266226 at n = 1, 36, ... .
The number of dual-Zeckendorf-infinitary divisors of n (defined in A331109) that are prime powers (A246655).
a(n) depends only on the prime signature of n.
Analogous to A064547 (binary representation) and A318464 (Zeckendorf representation).

Examples

			For n = 8 = 2^3, the dual Zeckendorf representation of 3 is 11, i.e., 3 = Fibonacci(2) + Fibonacci(3) = 1 + 2. Therefore 8 = 2^(1+2) = 2^1 * 2^2, and a(8) = 2.
For n = 256 = 2^8, the dual Zeckendorf representation of 8 is 1011, i.e., 8 = Fibonacci(2) + Fibonacci(3) + Fibonacci(5) = 1 + 2 + 5. Therefore 256 = 2^(1+2+5) = 2^1 * 2^2 * 2^5, and a(256) = 3.
		

Crossrefs

Programs

  • Mathematica
    toDualZeck[n_] := Module[{s = 0, v = 0, i = 0, f}, While[s < n, s += Fibonacci[i + 2]; v += 2^i; i++]; i--; While[i >= 0, f = Fibonacci[i + 2]; If[s - f >= n, s -= f; v -= 2^i]; i--]; v]; (* A003754, after Rémy Sigrist's PARI code in A112309 *)
    f[p_, e_] := DigitCount[toDualZeck[e], 2, 1]; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    todualzeck(n) = {my (s=0, v=0); for (i=0, oo, if (s>=n, forstep (j=i-1, 0, -1, if (s-fibonacci(2+j)>=n, s-=fibonacci(2+j); v-=2^j;);); return (v);); s+=fibonacci(2+i); v+=2^i;);} \\ A003754, Rémy Sigrist's code in A112309
    a(n) = vecsum(apply(x -> hammingweight(todualzeck(x)), factor(n)[, 2]));

Formula

Additive with a(p^e) = A112310(e).
a(n) = log_2(A331109(n)).
Sum_{k=1..n} a(k) ~ n * (log(log(n)) + B + C), where B is Mertens's constant (A077761), C = Sum_{k>=2} (A112310(k)-A112310(k-1)) * P(k) = 0.18790467121403662496..., and P(s) is the prime zeta function.
Previous Showing 41-50 of 59 results. Next